Question: The side of a square has the length $(x-2)$, while a rectangle has a length of $(x-3)$ and a width of $(x+4)$. If the area of the rectangle is twice the area of the square, what is the sum of the possible values of $x$?
Solution: The area of the square is $(x-2)^2$, while the area of the rectangle is $(x-3)(x+4)$. Now we set the area of the rectangle equal to twice the area of the square and solve for $x$: \begin{align*}
2(x-2)^2&=(x-3)(x+4)\quad\Rightarrow\\
2(x^2-4x+4)&=(x^2+x-12)\quad\Rightarrow\\
x^2-9x+20&=0\quad\Rightarrow\\
(x-5)(x-4)&=0.
\end{align*}So we know that $x=5$ or $x=4$, and our answer is $5+4=\boxed{9}$.